Find the value of \(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\)
Hint(s):What's the relationship between \(x\) and \(x+1\)?
Here is a computer summation of your continued fraction:
=1E100; listforeach(b,reverse(1,2,2,2.........etc ), c=b + 1/c);printc
c=1.4142135623730950488016887242097
The result is sqrt(2)
+118609
\(x = 1 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x = -1+2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2 + \ddots}}}}\\ x +1= 2 + \cfrac{1}{x+1}\\ \)
\(x +1= 2 + \cfrac{1}{x+1}\\ x-1= \cfrac{1}{x+1}\\ x^2-1=1\\ x^2=2\\ x=\pm\sqrt2 \)
I do not know why the answer cannot be -sqrt2 but if memory serves me correctly the only correct answer is +sqrt2
I'll continure thinking about that one.
